Integrand size = 14, antiderivative size = 74 \[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\frac {2 \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{4},\frac {6+m}{4},\sin ^2(a+b x)\right ) \sin (a+b x) \sqrt {c \sin ^m(a+b x)}}{b (2+m) \sqrt {\cos ^2(a+b x)}} \]
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Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3287, 2722} \[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\frac {2 \sin (a+b x) \cos (a+b x) \sqrt {c \sin ^m(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{4},\frac {m+6}{4},\sin ^2(a+b x)\right )}{b (m+2) \sqrt {\cos ^2(a+b x)}} \]
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Rule 2722
Rule 3287
Rubi steps \begin{align*} \text {integral}& = \left (\sin ^{-\frac {m}{2}}(a+b x) \sqrt {c \sin ^m(a+b x)}\right ) \int \sin ^{\frac {m}{2}}(a+b x) \, dx \\ & = \frac {2 \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{4},\frac {6+m}{4},\sin ^2(a+b x)\right ) \sin (a+b x) \sqrt {c \sin ^m(a+b x)}}{b (2+m) \sqrt {\cos ^2(a+b x)}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\frac {2 \sqrt {\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{4},\frac {6+m}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin ^m(a+b x)} \tan (a+b x)}{b (2+m)} \]
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\[\int \sqrt {c \left (\sin ^{m}\left (b x +a \right )\right )}d x\]
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Exception generated. \[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\int \sqrt {c \sin ^{m}{\left (a + b x \right )}}\, dx \]
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\[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\int { \sqrt {c \sin \left (b x + a\right )^{m}} \,d x } \]
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\[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\int { \sqrt {c \sin \left (b x + a\right )^{m}} \,d x } \]
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Timed out. \[ \int \sqrt {c \sin ^m(a+b x)} \, dx=\int \sqrt {c\,{\sin \left (a+b\,x\right )}^m} \,d x \]
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